@Chris My first though was to treat the sum in the numerator as a series and try to sum it up to, say - $k$ then bound given sequence from above by the sequence $\frac{k}{n}$ and from below by constant function $0$. But I can't see immediatly if it's the shortest way to evaluate that limit or if some extra difficulties wouldn't appear.
–
dataMay 25 '12 at 17:59

could you provide with some more details pls about that splitting. I'm surrounded by mist because maybe it's something i don't catch it yet. This is from my highschool notebook and i was thinking that there is an easier way to solve it. I'm struggling to understand your way, now.
–
Chris's sis the artistMay 25 '12 at 19:12

thanks for your solution. It's a bit ugly but i need to get used with these not-so-nice approaches.
–
Chris's sis the artistMay 25 '12 at 20:17

A more convenient way to state the sequence is:
$$
\frac{\sum_{k=1}^n\cos\frac{1}{k}\arccos\frac{1}{n-k+1}}{n}
$$

Note that for $0\leq x\leq 1$ we have $1-x\leq\cos x\leq 1$ and $\frac{\pi}{2}-\frac{\pi}{2}x\leq\arccos x\leq \frac{\pi}{2}-x$. Therefore, we have
$$
\tfrac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})\leq\cos\frac{1}{k}\arccos\frac{1}{n-k+1}\leq\frac{\pi}{2}-\frac{1}{n-k+1}.
$$
Thus, a lower bound for the limit is given by the sequence
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})}{n}=\frac{\pi}{2}-\frac{\sum_{k=0}^n\frac{(n-k+1)+k-1}{k(n-k+1)}}{n}=\frac{\pi}{2}-\sum_{k=1}^n\frac{1}{k(n-k+1)}
$$
The terms in the latter sum can be rewritten to
$$
\frac{1}{k(n+1)}+\frac{1}{(n-k+1)(n+1)}
$$
so we get the sums
$$
\frac{1}{n+1}\sum_{k=1}^n\frac{1}{k}\qquad\text{and}\qquad\frac{1}{n+1}\sum_{k=1}^n\frac{1}{n-k+1}.
$$
Sacha indicated a proof in the comments that these sums tend to zero. Similarly, for the upper bound we have
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}-\tfrac{1}{n-k+1}}{n}=\frac{\pi}{2}-\frac{\sum_{k=1}^n\frac{1}{n-k+1}}{n}\to\frac{\pi}{2}
$$
This gives an upper bound of $\frac{\pi}{2}$. This finishes the proof that the limit is $\frac{\pi}{2}$.

Note: Another proof that $\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\to 0$. Using Cauchy-Schwarz, we see that
$$
\sum_{k=1}^n\frac{1}{k}\leq \sqrt{n}\sqrt{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}
$$
Therefore we get
$$
\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\leq\sqrt{\frac{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}{n}}
$$
In the square root on the right hand side, the numerator converges, so the whole tends to zero.

+1 This is exactly the way to go. You can now show that $\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{n+1-k} = \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k} = \lim_{n\to \infty} \frac{1}{n} H_n = 0$, and similarly for the sum of the lower bound.
–
SashaMay 25 '12 at 19:42

This is becoming a community proof now. Thanks @Sasha!
–
EgbertMay 25 '12 at 20:34

As Chris notes in the comment below this answer, it is possible to prove that $\frac{\pi}2$ is also a lower bound by a simple application of AM-GM inequality and the Cesaro-Stolz theorem, completing the proof.

@Chris: That's a great idea! There may be some complications caused by the fact that $\arccos1=0$ but we can simply throw that last term away and write $\frac1n=\frac1{n-1}\frac{n-1}n$, allowing us to use AM-GM on the remaining $n-1$ terms, which are positive.
–
Dejan GovcMay 26 '12 at 17:52